Copyright © 2011 Pearson, Inc. 6.1 Vectors in the Plane

Copyright © 2011 Pearson, Inc.

6.1Vectors in the

Plane

Slide 6.1 - 2 Copyright © 2011 Pearson, Inc.

What you’ll learn about

Two-Dimensional Vectors Vector Operations Unit Vectors Direction Angles Applications of Vectors

… and whyThese topics are important in many real-world applications, such as calculating the effect of the wind on an airplane’s path.


Directed Line Segment


Two-Dimensional Vector

A two - dimensional vector v is an ordered pair of real

numbers, denoted in component form as a,b . The

numbers a and b are the components of the vector v.

The standard representation of the vector a,b is the

arrow from the origin to the point (a,b). The magnitude

of v is the length of the arrow and the direction of v is the

direction in which the arrow is pointing. The vector

0 = 0,0 , called the zero vector, has zero length and

no direction.


Initial Point, Terminal Point, Equivalent


Head Minus Tail (HMT) Rule

If an arrow has initial point x1, y

1( ) and terminal point

x2 , y2( ), it represents the vector x2 −x1, y2 −y1 .


Magnitude

If v is represented by the arrow from x1, y

1( ) to x2 , y2( ),

then

v = x2 −x1( )2+ y2 −y1( )

2.

If v= a,b , then v = a2 +b2 .


Example Finding Magnitude of a Vector

Find the magnitude of v represented by PQ,

where P =(3,−4) and Q=(5,2).


Example Finding Magnitude of a Vector

v = x2 −x1( )2+ y2 −y1( )

2

= 5−3( )2+ 2−(−4)( )

2

=2 10

Find the magnitude of v represented by PQ,

where P =(3,−4) and Q=(5,2).


Vector Addition and Scalar Multiplication

Let u = u1,u2 and v= v1,v2 be vectors and let k be

a real number (scalar). The sum (or resultant) of the

vectors u and v is the vector

u+ v= u1 +v1,u2 +v2 .

The product of the scalar k and the vector u is

ku=k u1,u2 = ku1,ku2 .


Example Performing Vector Operations

Let u = 2, −1 and v= 5,3 . Find 3u+ v.


Example Performing Vector Operations

3u = 3 2( ), 3 −1( ) = 6,−3

3u+ v = 6,−3 + 5,3 = 6+5,−3+3 = 11,0

Let u = 2, −1 and v= 5,3 . Find 3u+ v.


Unit Vectors

A vector u with | u | =1 is a unit vector. If v is not

the zero vector 0,0 , then the vector u=v|v |

=1|v |

v

is a unit vector in the direction of v.


Example Finding a Unit Vector

Find a unit vector in the direction of v = 2, −3 .


Example Finding a Unit Vector

v = 2, −3 = 22 + −3( )2= 13, so

v

v=

1

132, −3 =

2

13, −

3

13

Find a unit vector in the direction of v = 2, −3 .


Standard Unit Vectors

The two vectors i = 1,0 and j = 0,1 are the standard

unit vectors. Any vector v can be written as an expression

in terms of the standard unit vector:

v= a,b

= a,0 + 0,b

=a 1,0 +b 0,1

=ai +bj


Resolving the Vector

If v has direction angle θ, the components of v canbe computed using the formula

v = vcosθ, vsinθ .

From the formula above, it follows that the unit vector

in the direction of v is u =vv= cosθ,sinθ .


Example Finding the Components of a Vector

Find the components of the vector v with

direction angle 120o and magnitude 8.


Example Finding the Components of a Vector

v = a,b = 8cos120o,8sin120o

= 8 −12

⎛

⎝⎜⎞

⎠⎟,8

32

⎛

⎝⎜

⎞

⎠⎟

= −4,4 3

So a=−4 and b=4 3.

Find the components of the vector v with

direction angle 120o and magnitude 8.


Example Finding the Direction Angle of a Vector

Let u = −2,3 and v= −4,−1 .

Find the magnitude and direction angle of each vector.



Let α be the direction angle of u, then

|u|= −2( )2+32 = 13

−2 =u cosα

−2 = 13cosα

cosα =−2

13

Let u = −2,3 and v= −4,−1 .


α =cos−1−2

13

⎛

⎝⎜⎞

⎠⎟

90º<α <180º

α ≈123.69º



Let β be the direction angle of v, then

|v |= −4( )2+ −1( )

2= 17

−4 = v cosβ

−4 = 17 cosβ

cosβ =−4

17

Let u = −2,3 and v= −4,−1 .


β =360º−cos−1−4

17

⎛

⎝⎜⎞

⎠⎟

180º< β < 2700º

β ≈194.04º



Interpret

The direction angle for u

is α ≈123.69º.

The direction angle for vis α ≈194.04º.

Let u = −2,3 and v= −4,−1 .



Velocity and Speed

The velocity of a moving object is a vector

because velocity has both magnitude and

direction. The magnitude of velocity is speed.


Quick Review

1. Find the values of x and y.

2. Solve for θ in degrees. θ =sin-1 3

11

⎛

⎝⎜⎞

⎠⎟


Quick Review

3. A naval ship leaves Port Northfolk and averages

43 knots (nautical mph) traveling for 3 hr on a bearing

of 35o and then 4 hr on a course of 120o.

What is the boat's bearing and distance from

Port Norfolk after 7 hr.

The point P is on the terminal side of the angle θ. Find the measure of θ if 0o <θ < 360o.4. P(5,7)5. P(-5,7)


Quick Review Solutions

1. Find the values of x and y.

x =−7.5, y=7.5 3

2. Solve for θ in degrees. θ =sin-1 3

11

⎛

⎝⎜⎞

⎠⎟64.8º


Quick Review Solutions

3. A naval ship leaves Port Northfolk and averages

43 knots (nautical mph) traveling for 3 hr on a bearing

of 35o and then 4 hr on a course of 120o.

What is the boat's bearing and distance from

Port Norfolk after 7 hr.

distance = 224.2; bearing = 84.9º

The point P is on the terminal side of the angle θ. Find the measure of θ if 0o <θ < 360o.4. P(5,7) 54.5º5. P(−5,7) 125.5º

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Copyright © 2011 Pearson, Inc. 6.1 Vectors in the Plane